Simplify the following expression and state the condition under which the simplification is valid. You can assume that $z \neq 0$. $a = \dfrac{7(5z + 8)}{4z} \div \dfrac{50z + 80}{6z} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{7(5z + 8)}{4z} \times \dfrac{6z}{50z + 80} $ When multiplying fractions, we multiply the numerators and the denominators. $a = \dfrac{ 7(5z + 8) \times 6z } { 4z \times (50z + 80) } $ $ a = \dfrac {6z \times 7(5z + 8)} {4z \times 10(5z + 8)} $ $ a = \dfrac{42z(5z + 8)}{40z(5z + 8)} $ We can cancel the $5z + 8$ so long as $5z + 8 \neq 0$ Therefore $z \neq -\dfrac{8}{5}$ $a = \dfrac{42z \cancel{(5z + 8})}{40z \cancel{(5z + 8)}} = \dfrac{42z}{40z} = \dfrac{21}{20} $